Let $d\in\mathbb N$, $M\subseteq\mathbb R^d$, $f:M\to\mathbb R$ and $\alpha\in\mathbb N_0$.
If $M$ is not open, we usually mean by saying that "$f$ is $C^\alpha$-differentiable" that there is an open neighborhood $\Omega$ of $M$ and a function $\overline f\in C^\alpha(\Omega)$ with $\left.\overline f\right|_M=f$.
On the other hand, in the context of differential geometry, if $M$ is a $k$-dimensional embedded $C^\alpha$-submanifold, we usually write $f\in C^\alpha(M)$ if $$f\circ\phi^{-1}\in C^\alpha(U)\tag1$$ for each $C^\alpha$-diffeomorphism$^1$ $\phi$ from an open subset of $M$ onto an open subset $U$ of $\mathbb R^k$.
So, at first glance, assuming the latter case, it seems like writing "$f\in C^\alpha(M)$" is ambiguous, unless we can show that both conditions are equivalent (or at least one of the implications hold). Can we show that?
$^1$ Since the domain $N$ of $\phi$ is only required to be open in $M$, $N$ doesn't need to open in $\mathbb R^d$. So, the claim $\phi\in C^\alpha(N,\mathbb R^k)$ has again to be understood in aforementioned sense. On the other hand, since $N$ is open in $M$, $N=V\cap M$ for some open subset $V$ of $\mathbb R^d$. But I'm unsure whether we can always choose $V$ such that $\phi$ is the restriction of a $C^\alpha(V,\mathbb R^d)$-function. Please note that I've asked a related question: Do diffeomorphisms between arbitrary sets extend to ordinary diffeomorphisms between open sets?.